Geo. Applications of Calculus (1 Viewer)

[Avenger6]

Hi, just wondering if anyone can help me solve this question regarding the practical use of calculus.

A 5m length of timber is used to border a triangular garden bed with the other sides of the garden against the house walls.

i.e.
___|\
___|=\
___|==\ 5m
__x|===\
___|====\
___|=====\
___-----y-----

If the area of the garden is given by the equation A=1/2x[(SQUARE ROOT)25-x²] find the greatest possible area of the garden bed.

Any help is greatly appreciated .

 

[Mark576]

Differentiate the given equation for the area and solve dA/dx = 0 to find any maximum, minimum values of x. Use either the first derivative method or the second derivative to determine whether these values give a max. area or min. area, and substitute the maximum value into the equation for A to find the maximum area.

 

[YannY]

A=1/2x[(SQUARE ROOT)25-x<SUP>2</SUP>]

let u=SQUARE ROOT(25-x^2)
=(25-x^2)^(1/2)
u'= -2x(25-x^2)^(-1/2)

so A=1/2 . x . u
A'=1/2u + 1/2 x u'
=1/2(25-x^2)^(1/2) + 1/2x . -2x(25-x^2)^(-1/2)
=(25-x^2)^(-1/2) . [1/2(25-x^2) - 1/ 2 . x^2]
let A'=0
i.e 25/2 - x^2 = 0
x= 5/rt2 and -5/rt2

therefore there is a maximum at x=5/rt2 and from pythagoros y=5/rt2 as well.

 

[YannY]

are is (5/rt2)^2 times a half i.e 25/4 m squared.

 

[namburger]

Hi, just wondering if anyone can help me solve this question regarding the practical use of calculus.

A 5m length of timber is used to border a triangular garden bed with the other sides of the garden against the house walls.

i.e.
___|\
___|=\
___|==\ 5m
__x|===\
___|====\
___|=====\
___-----y-----

If the area of the garden is given by the equation A=1/2x[(SQUARE ROOT)25-x²] find the greatest possible area of the garden bed.

Any help is greatly appreciated .

A = 0.5X(25-X^2)^0.5
Diffrentiate it using the product rule
A' = 0.5 [(25-X^2)^0.5 + 0.5(25-X^2)^-0.5 x -2X x X]
= 0.5 [(25-X^2)^0.5 - X^2(25-X^2)^-0.5]
Let A' = 0
[(25-X^2)^0.5 - X^2(25-X^2)^-0.5] = 0
25- X^2 - X^2 = 0
X^2 = 12.5
X = 3.53

Sub that into
A = 0.5X(25-X^2)^0.5
= 6.25m^2

Sorta messy, remember to prove it is a max at X = 3.53

 

[Avenger6]

Im sorry guys, I've poorly written the formula. I have uploaded a more accurate copy, the x is actually a pronumeral not a multiplaction sign. Im sorry for wasting your time, I should have made it more clear. Any help would be greatly appreciated however. Once again, my apologies.

 

[YannY]

i didnt regard your x as a multiplication sign. Nor did namburger.

 

[Avenger6]

Ok thanks guys, made sense. Just one question, I see how you guys found the stationary point but how did you prove it was a maximum point, don't you need the second derivative for that? I am also a little lost as to what happened with the 0.5 out the front...

 

[Mark576]

If F''(x)>0, then the point at x is a minimum.
If F''(x)<0, then the point at x is a maximum.
OR we can use the first derivative:
If F'(x) changes from positive to negative through x, then (x,F(x)) is a maximum stationary point, conversely, if it changes from negative to positive, then (x,F(x)) is a minimum.

 

[namburger]

Ok thanks guys, made sense. Just one question, I see how you guys found the stationary point but how did you prove it was a maximum point, don't you need the second derivative for that?

There are 3 methods i kno:
1. Find second derivative.

2. Sub points very close to the maximum. i.e. x= 3.52 and 3.54 into the first derivative equation and show that when x =3.52, y' = postiive and when x =3.54, y' = negative

3. Finding the sign of y'. This works by removing positive values from the y' equation and drawing the y' curve. you will see that to the left of x=3.53 it is positive and to the right it is negative.

I use 3. but its kinda difficult to explain.

 

[YannY]

it should be obvious that it is a maximum since the Area cannot have a minimum unless it goes to negative values or 0. But area cannot be negative so we disregard those that are negative. So now we have a stationary point that is a positive - do you think it would be a minimum value?

 

[FDownes]

Wow, I just came online to ask the exact same question! That's handy.